HSC 2013 MX2 Marathon (archive) (2 Viewers)

HeroicPandas · Dec 17, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
That is one case of it, yes.

I should of extended the question to:

$Find the possible roots of S(x)$

So yeah consider if -1 is the first or the last term etc.

When x=-1 is first root

roots: -1, -1+d, -1+2d

$\prod \alpha:$ -1 (d-1)(2d-1) = -1

2d^2 - 3d=0
d(2d-3) =0
d=0 (done that) d= 3/2

therefore roots: -1, -1 +3/2, -1 +3

When x=-1 is last root

roots: -1- 2d, -1 -d, -1

$\prod \alpha:$ (2d+1)(d+1)=1
2d^2 +3d=0
d(2d+3)=0
d=0 (done that) d=-3/2

therefore roots: -1 +3, -1 + 3/2, -1

but then...roots do not satisfy equation S(x)

Sy123 · Dec 17, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
When x=-1 is first root

roots: -1, -1+d, -1+2d

$\prod \alpha:$ -1 (d-1)(2d-1) = -1

2d^2 - 3d=0
d(2d-3) =0
d=0 (done that) d= 3/2

therefore roots: -1, -1 +3/2, -1 +3

When x=-1 is last root

roots: -1- 2d, -1 -d, -1

$\prod \alpha:$ (2d+1)(d+1)=1
2d^2 +3d=0
d(2d+3)=0
d=0 (done that) d=-3/2

therefore roots: -1 +3, -1 + 3/2, -1

but then...roots do not satisfy equation S(x)

Yes this is what I intended but it seems that the roots do not work....

Can a math pro please explain why this is the case?

HeroicPandas · Dec 17, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Yes this is what I intended but it seems that the roots do not work....

Can a math pro please explain why this is the case?

OOO OOO yes, I want to know too

Carrotsticks · Dec 17, 2012

Re: HSC 2013 4U Marathon

This is because we have a Symmetric Polynomial (sometimes also called a Reciprocal Polynomial due to nature of the roots), and a distinguishing feature of them is that if x is a root, then 1/x is also a root. This is easily proven by finding P(1/x).

So the three possible cases you have there ARE correct, except only one of them satisfy the rule that if x is a root, then 1/x is also a root.

The only one that satisfies this, is when the three roots are x=-1 -1 and -1 (ie: triple root). It is in arithmetic progression, in the sense that 0 is the common difference and -1 is the starting value.

seanieg89 · Dec 17, 2012

Re: HSC 2013 4U Marathon

And it shouldn't be surprising. You have used information about the poly. to give you an equation d must satisfy. This does NOT mean that every solution of this equation is a valid value for d.

This is an example of "if" vs "only if".

(And also an illustration of how it is often nicer to make symmetric choices: ie a-d,a,a+d vs a,a+d,a+2d.)

Sy123 · Dec 17, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
This is because we have a Symmetric Polynomial (sometimes also called a Reciprocal Polynomial due to nature of the roots), and a distinguishing feature of them is that if x is a root, then 1/x is also a root. This is easily proven by finding P(1/x).

So the three possible cases you have there ARE correct, except only one of them satisfy the rule that if x is a root, then 1/x is also a root.

The only one that satisfies this, is when the three roots are x=-1 -1 and -1 (ie: triple root). It is in arithmetic progression, in the sense that 0 is the common difference and -1 is the starting value.

seanieg89 said:
And it shouldn't be surprising. You have used information about the poly. to give you an equation d must satisfy. This does NOT mean that every solution of this equation is a valid value for d.

This is an example of "if" vs "only if".

(And also an illustration of how it is often nicer to make symmetric choices: ie a-d,a,a+d vs a,a+d,a+2d.)

Alright thanks guys

RealiseNothing · Dec 18, 2012

Re: HSC 2013 4U Marathon

Also note it is a cubic, hence it can only have 3 roots. If we were to take all cases, we would end up with many more roots, so only some will work.

HeroicPandas · Dec 18, 2012

Re: HSC 2013 4U Marathon

LET m BE A POSITIVE INTEGER

$$a) By using de Moivre's theorem, show that: $\\ sin(2m+1)\theta = \binom{2m+1}{1}cos^{2m}\theta sin\theta - \binom{2m+1}{3}cos^{2m-2}\theta sin^{3}\theta +...+ (-1)^{m}sin^{2m+1}\theta$

$$b) Deduce that the polynomial\\ P(x) \equiv \binom{2m+1}{1}x^m - \binom{2m+1}{3}x^{m-1}+...+(-1)^m$

$$has m distinct roots$\\ cot^2(\frac{k\pi}{2m+1}) where$ k=1, 2,...,m$

$$c) Prove that$ \\ cot^2(\frac{\pi}{2m+1}) + cot^2(\frac{2\pi}{2m+1})+...+cot^2(\frac{m\pi}{2m+1})=\frac{m(2m-1)}{}3$

$$d) You are given that cot\theta < \frac{1}{\theta}$

^FOR 0< $\theta$ < $\frac{\pi}{2}$

DEDUCE THAT:

$\frac{\pi^2}{6}<(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{m^2}) \frac{(2m+1)^2}{2m(2m-1)}$

I will post up my solution for part d) if someone uses a different method compared to mine

Sy123 · Dec 18, 2012

Re: HSC 2013 4U Marathon

Heroic your reasoning for the why it is (-4)^n is wrong but all up until the last part is right. Rethink what it means whether n is odd or even.

I will do your question now.

Sy123 · Dec 18, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
LET m BE A POSITIVE INTEGER

$$a) By using de Moivre's theorem, show that: $\\ sin(2m+1)\theta = \binom{2m+1}{1}cos^{2m}\theta sin\theta - \binom{2m+1}{3}cos^{2m-2}\theta sin^{3}\theta +...+ (-1)^{m}sin^{2m+1}\theta$

$$b) Deduce that the polynomial\\ P(x) \equiv \binom{2m+1}{1}x^m - \binom{2m+1}{3}x^{m-1}+...+(-1)^m$

$$has m distinct roots$\\ cot^2(\frac{k\pi}{2m+1}) where$ k=1, 2,...,m$

$$c) Prove that$ \\ cot^2(\frac{\pi}{2m+1}) + cot^2(\frac{2\pi}{2m+1})+...+cot^2(\frac{m\pi}{2m+1})=\frac{m(2m-1)}{}3$

$$d) You are given that cot\theta < \frac{1}{\theta}$

^FOR 0< $\theta$ < $\frac{\pi}{2}$

DEDUCE THAT:

$\frac{\pi^2}{6}<(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{m^2}) \frac{(2m+1)^2}{2m(2m-1)}$

I will post up my solution for part d) if someone uses a different method compared to mine

Good question, I'll type necessary details:

a)

$(\cos \theta +i\sin \theta)^{2m+1} = \binom{2m+1}{0}\cos^{2m+1} \theta +...$

Equate imaginary parts and the solution falls out. We get the last term as (-1)^m as i^(odd) will either be -i or -1. And since we equate imaginary parts we disregard -i.

b)

$x=\frac{\cos^2 \theta}{\sin^2 \theta}$

Sub in P(x). AND multiply both sides by $\sin^{2m+1} \theta$

We end up with:

$P(x)= \frac{\sin(2m+1)\theta}{\sin^{2m+1}\theta}$

In order for P(x) to be zero, $\sin(2m+1)\theta = 0$
And we arrive at:

$\theta = \frac{k\pi}{2m+1}$

Now sub that back into x it self, notice how I made x= cot squared theta.

So when we sub it back in, we arrive at the result we are trying to prove. k is valid only for 1, 2, 3 ... m because there are m roots.

c)

Simply do sum of all roots:

$\sum_{k=1}^m \cot^2 \left (\frac{k\pi}{2m+1})\right = \binom{2m+1}{3} \div \binom{2m+1}{1}$

Which after simplification yields the given result.

d)

We are given that theta is positive, hence we can produce the argument that:

$\cot^2 \theta < 1/\theta^2$

Let theta be the respective term for each sequence. We arrive at the result that (after simplification). We are justified in shifting since we only shift the positive.

$\frac{(2m+1)^2}{1^2\pi^2}+\frac{(2m+1)^2}{2^2\pi^2}+...+\frac{(2m+1)^2)}{m^2 \pi^2}>\frac{m}{3}(2m-1)$

Take (2m+1)^2 / pi^2 common, then shift stuff to respective sides, then multiply whole thing by 1/2

We yield the required result.

$\frac{(2m+1)^2}{2m(2m-1)} \cdot \sum_{k=1}^m \frac{1}{k^2} > \frac{\pi^2}{6}$
$\square$

====

HeroicPandas · Dec 18, 2012

Re: HSC 2013 4U Marathon

WOAH! NICE! GOOD WORK!

$cot^{2}\theta <1/\theta^2$

What i did was, i let $\theta = \frac{k\pi}{2m+1}$

So, $cot^{2}(\frac{k\pi}{2m+1}) < \frac{(2m+1)^2}{k^2 \pi^2}$

$\sum_{k=1}^{m}$ both sides

$\frac{m(2m-1)}{3} < \frac{(2m+1)^2}{\pi^2} \sum_{k=1}^{m}\frac{1}{k^2}$

RE-arrange times both sides by $\frac{1}{2}$

I skipped many steps to save time

OK- can u explain the part when u wrote AFTER $cot^2 \theta < 1/\theta^2$

"Let theta be the respective term for each sequence. We arrive at the result after simplification. We are justified in shifting since we only shift the positive."

Sy123 · Dec 18, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
WOAH! NICE!

$cot^{2}\theta <1/\theta^2$

What i did was, i let $\theta = \frac{k\pi}{2m+1}$

So, $cot^{2}(\frac{k\pi}{2m+1}) < \frac{(2m+1)^2}{k^2 \pi^2}$

$\sum_{k=1}^{m}$ both sides

$\frac{m(2m-1)}{3} < \frac{(2m+1)^2}{\pi^2} \sum_{k=1}^{m}1/k^2$

RE-arrange times both sides by $\frac{1}{2}$

OK- can u explain the part when u wrote $cot^2 \theta < 1/\theta^2$

"Let theta be the respective term for each sequence. We arrive at the result after simplification. We are justified in shifting since we only shift the positive."

I cant be bothered writing the whole sequence out for each theta, so I will do this:

Let

$\theta_k= \frac{k\pi}{2m+1}$

$\cot^2 \theta_1 + \cot^2 \theta_2+...+\cot^2 \theta_m= $something$$

$\cot^2 \theta_k < 1/\theta_k^2$

$1/\theta_1^2+1/\theta_2^2+...+1/\theta_m^2 > $something$$

Basically the same thing you did

GoldyOrNugget · Dec 18, 2012

Re: HSC 2013 4U Marathon

Wasn't that part of the CSSA 2012 Q16? Or the final question of a past HSC paper? It looks very familiar.

HeroicPandas · Dec 18, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Wasn't that part of the CSSA 2012 Q16? Or the final question of a past HSC paper? It looks very familiar.

I don't know lol, i got it from tutoring handout

EDIT: if it is and u have the answers, can u post it up? i wanna see what they did

HeroicPandas · Dec 18, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Basically the same thing you did

Oh ok

GoldyOrNugget · Dec 18, 2012

Re: HSC 2013 4U Marathon

Hmm so the CSSA q16 was different, but the 'Deduce that' was in there, as the third last part or something.

bleakarcher · Dec 18, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Hmm so the CSSA q16 was different, but the 'Deduce that' was in there, as the third last part or something.

The exact question is found in Q7 of a past HSC paper from memory, 2002/3 maybe.

HeroicPandas · Dec 18, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
The exact question is found in Q7 of a past HSC paper from memory, 2002/3 maybe.

Yeh its the 2002 HSC paper, i just checked (Question 8)

Sy123 · Dec 18, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Wasn't that part of the CSSA 2012 Q16? Or the final question of a past HSC paper? It looks very familiar.

Nope, CSSA Q16 was a real finding the Basel Problem rather than just the expression in this case.

Though for the result at the end of Heroic's question, I have a question for the maths pros

If I take the limit to infinity of the RHS of Heroic's last result. I get the sum of squares of the reciprocals of the natural numbers (basel problem) (since the fraction on the very right converges to 1). How would I prove that at m infinite the inequality becomes equality?

bleakarcher · Dec 18, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
Yeh its the 2002 HSC paper, i just checked (Question 8)

Oh yeah right, q8 sorry.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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